3.21 \(\int x (d+e x)^3 (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=122 \[ -\frac{1}{20} \left (\frac{5 d (d+e x)^4}{e^2}-\frac{4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac{b d^5 n \log (x)}{20 e^2}+\frac{1}{15} b d^2 e n x^3+\frac{b d^4 n x}{5 e}+\frac{3}{20} b d^3 n x^2+\frac{1}{80} b d e^2 n x^4-\frac{b n (d+e x)^5}{25 e^2} \]

[Out]

(b*d^4*n*x)/(5*e) + (3*b*d^3*n*x^2)/20 + (b*d^2*e*n*x^3)/15 + (b*d*e^2*n*x^4)/80 - (b*n*(d + e*x)^5)/(25*e^2)
+ (b*d^5*n*Log[x])/(20*e^2) - (((5*d*(d + e*x)^4)/e^2 - (4*(d + e*x)^5)/e^2)*(a + b*Log[c*x^n]))/20

________________________________________________________________________________________

Rubi [A]  time = 0.0911198, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {43, 2334, 12, 80} \[ -\frac{1}{20} \left (\frac{5 d (d+e x)^4}{e^2}-\frac{4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac{b d^5 n \log (x)}{20 e^2}+\frac{1}{15} b d^2 e n x^3+\frac{b d^4 n x}{5 e}+\frac{3}{20} b d^3 n x^2+\frac{1}{80} b d e^2 n x^4-\frac{b n (d+e x)^5}{25 e^2} \]

Antiderivative was successfully verified.

[In]

Int[x*(d + e*x)^3*(a + b*Log[c*x^n]),x]

[Out]

(b*d^4*n*x)/(5*e) + (3*b*d^3*n*x^2)/20 + (b*d^2*e*n*x^3)/15 + (b*d*e^2*n*x^4)/80 - (b*n*(d + e*x)^5)/(25*e^2)
+ (b*d^5*n*Log[x])/(20*e^2) - (((5*d*(d + e*x)^4)/e^2 - (4*(d + e*x)^5)/e^2)*(a + b*Log[c*x^n]))/20

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rubi steps

\begin{align*} \int x (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac{1}{20} \left (\frac{5 d (d+e x)^4}{e^2}-\frac{4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac{(d+e x)^4 (-d+4 e x)}{20 e^2 x} \, dx\\ &=-\frac{1}{20} \left (\frac{5 d (d+e x)^4}{e^2}-\frac{4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{(b n) \int \frac{(d+e x)^4 (-d+4 e x)}{x} \, dx}{20 e^2}\\ &=-\frac{b n (d+e x)^5}{25 e^2}-\frac{1}{20} \left (\frac{5 d (d+e x)^4}{e^2}-\frac{4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac{(b d n) \int \frac{(d+e x)^4}{x} \, dx}{20 e^2}\\ &=-\frac{b n (d+e x)^5}{25 e^2}-\frac{1}{20} \left (\frac{5 d (d+e x)^4}{e^2}-\frac{4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac{(b d n) \int \left (4 d^3 e+\frac{d^4}{x}+6 d^2 e^2 x+4 d e^3 x^2+e^4 x^3\right ) \, dx}{20 e^2}\\ &=\frac{b d^4 n x}{5 e}+\frac{3}{20} b d^3 n x^2+\frac{1}{15} b d^2 e n x^3+\frac{1}{80} b d e^2 n x^4-\frac{b n (d+e x)^5}{25 e^2}+\frac{b d^5 n \log (x)}{20 e^2}-\frac{1}{20} \left (\frac{5 d (d+e x)^4}{e^2}-\frac{4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.118846, size = 130, normalized size = 1.07 \[ d^2 e x^3 \left (a+b \log \left (c x^n\right )\right )+\frac{1}{2} d^3 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac{3}{4} d e^2 x^4 \left (a+b \log \left (c x^n\right )\right )+\frac{1}{5} e^3 x^5 \left (a+b \log \left (c x^n\right )\right )-\frac{1}{3} b d^2 e n x^3-\frac{1}{4} b d^3 n x^2-\frac{3}{16} b d e^2 n x^4-\frac{1}{25} b e^3 n x^5 \]

Antiderivative was successfully verified.

[In]

Integrate[x*(d + e*x)^3*(a + b*Log[c*x^n]),x]

[Out]

-(b*d^3*n*x^2)/4 - (b*d^2*e*n*x^3)/3 - (3*b*d*e^2*n*x^4)/16 - (b*e^3*n*x^5)/25 + (d^3*x^2*(a + b*Log[c*x^n]))/
2 + d^2*e*x^3*(a + b*Log[c*x^n]) + (3*d*e^2*x^4*(a + b*Log[c*x^n]))/4 + (e^3*x^5*(a + b*Log[c*x^n]))/5

________________________________________________________________________________________

Maple [C]  time = 0.225, size = 598, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(e*x+d)^3*(a+b*ln(c*x^n)),x)

[Out]

ln(c)*b*d^2*e*x^3+3/4*ln(c)*b*d*e^2*x^4+1/2*a*d^3*x^2+1/4*I*Pi*b*d^3*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+1/4*I*Pi*
b*d^3*x^2*csgn(I*c*x^n)^2*csgn(I*c)+3/4*a*d*e^2*x^4+a*d^2*e*x^3-1/4*I*Pi*b*d^3*x^2*csgn(I*c*x^n)^3-1/10*I*Pi*b
*e^3*x^5*csgn(I*c*x^n)^3+1/2*I*Pi*b*d^2*e*x^3*csgn(I*c*x^n)^2*csgn(I*c)+1/2*I*Pi*b*d^2*e*x^3*csgn(I*x^n)*csgn(
I*c*x^n)^2-1/10*I*Pi*b*e^3*x^5*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/4*I*Pi*b*d^3*x^2*csgn(I*x^n)*csgn(I*c*x^n
)*csgn(I*c)+3/8*I*Pi*b*d*e^2*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2+3/8*I*Pi*b*d*e^2*x^4*csgn(I*c*x^n)^2*csgn(I*c)-3/
8*I*Pi*b*d*e^2*x^4*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/2*I*Pi*b*d^2*e*x^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c
)-1/2*I*Pi*b*d^2*e*x^3*csgn(I*c*x^n)^3+1/10*I*Pi*b*e^3*x^5*csgn(I*x^n)*csgn(I*c*x^n)^2+1/10*I*Pi*b*e^3*x^5*csg
n(I*c*x^n)^2*csgn(I*c)+1/5*a*e^3*x^5-3/8*I*Pi*b*d*e^2*x^4*csgn(I*c*x^n)^3+1/5*ln(c)*b*e^3*x^5+1/2*ln(c)*b*d^3*
x^2+1/20*b*x^2*(4*e^3*x^3+15*d*e^2*x^2+20*d^2*e*x+10*d^3)*ln(x^n)-1/4*b*d^3*n*x^2-1/3*b*d^2*e*n*x^3-3/16*b*d*e
^2*n*x^4-1/25*b*e^3*n*x^5

________________________________________________________________________________________

Maxima [A]  time = 0.984495, size = 190, normalized size = 1.56 \begin{align*} -\frac{1}{25} \, b e^{3} n x^{5} + \frac{1}{5} \, b e^{3} x^{5} \log \left (c x^{n}\right ) - \frac{3}{16} \, b d e^{2} n x^{4} + \frac{1}{5} \, a e^{3} x^{5} + \frac{3}{4} \, b d e^{2} x^{4} \log \left (c x^{n}\right ) - \frac{1}{3} \, b d^{2} e n x^{3} + \frac{3}{4} \, a d e^{2} x^{4} + b d^{2} e x^{3} \log \left (c x^{n}\right ) - \frac{1}{4} \, b d^{3} n x^{2} + a d^{2} e x^{3} + \frac{1}{2} \, b d^{3} x^{2} \log \left (c x^{n}\right ) + \frac{1}{2} \, a d^{3} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)^3*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

-1/25*b*e^3*n*x^5 + 1/5*b*e^3*x^5*log(c*x^n) - 3/16*b*d*e^2*n*x^4 + 1/5*a*e^3*x^5 + 3/4*b*d*e^2*x^4*log(c*x^n)
- 1/3*b*d^2*e*n*x^3 + 3/4*a*d*e^2*x^4 + b*d^2*e*x^3*log(c*x^n) - 1/4*b*d^3*n*x^2 + a*d^2*e*x^3 + 1/2*b*d^3*x^
2*log(c*x^n) + 1/2*a*d^3*x^2

________________________________________________________________________________________

Fricas [A]  time = 1.02135, size = 394, normalized size = 3.23 \begin{align*} -\frac{1}{25} \,{\left (b e^{3} n - 5 \, a e^{3}\right )} x^{5} - \frac{3}{16} \,{\left (b d e^{2} n - 4 \, a d e^{2}\right )} x^{4} - \frac{1}{3} \,{\left (b d^{2} e n - 3 \, a d^{2} e\right )} x^{3} - \frac{1}{4} \,{\left (b d^{3} n - 2 \, a d^{3}\right )} x^{2} + \frac{1}{20} \,{\left (4 \, b e^{3} x^{5} + 15 \, b d e^{2} x^{4} + 20 \, b d^{2} e x^{3} + 10 \, b d^{3} x^{2}\right )} \log \left (c\right ) + \frac{1}{20} \,{\left (4 \, b e^{3} n x^{5} + 15 \, b d e^{2} n x^{4} + 20 \, b d^{2} e n x^{3} + 10 \, b d^{3} n x^{2}\right )} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)^3*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

-1/25*(b*e^3*n - 5*a*e^3)*x^5 - 3/16*(b*d*e^2*n - 4*a*d*e^2)*x^4 - 1/3*(b*d^2*e*n - 3*a*d^2*e)*x^3 - 1/4*(b*d^
3*n - 2*a*d^3)*x^2 + 1/20*(4*b*e^3*x^5 + 15*b*d*e^2*x^4 + 20*b*d^2*e*x^3 + 10*b*d^3*x^2)*log(c) + 1/20*(4*b*e^
3*n*x^5 + 15*b*d*e^2*n*x^4 + 20*b*d^2*e*n*x^3 + 10*b*d^3*n*x^2)*log(x)

________________________________________________________________________________________

Sympy [A]  time = 3.67076, size = 218, normalized size = 1.79 \begin{align*} \frac{a d^{3} x^{2}}{2} + a d^{2} e x^{3} + \frac{3 a d e^{2} x^{4}}{4} + \frac{a e^{3} x^{5}}{5} + \frac{b d^{3} n x^{2} \log{\left (x \right )}}{2} - \frac{b d^{3} n x^{2}}{4} + \frac{b d^{3} x^{2} \log{\left (c \right )}}{2} + b d^{2} e n x^{3} \log{\left (x \right )} - \frac{b d^{2} e n x^{3}}{3} + b d^{2} e x^{3} \log{\left (c \right )} + \frac{3 b d e^{2} n x^{4} \log{\left (x \right )}}{4} - \frac{3 b d e^{2} n x^{4}}{16} + \frac{3 b d e^{2} x^{4} \log{\left (c \right )}}{4} + \frac{b e^{3} n x^{5} \log{\left (x \right )}}{5} - \frac{b e^{3} n x^{5}}{25} + \frac{b e^{3} x^{5} \log{\left (c \right )}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)**3*(a+b*ln(c*x**n)),x)

[Out]

a*d**3*x**2/2 + a*d**2*e*x**3 + 3*a*d*e**2*x**4/4 + a*e**3*x**5/5 + b*d**3*n*x**2*log(x)/2 - b*d**3*n*x**2/4 +
b*d**3*x**2*log(c)/2 + b*d**2*e*n*x**3*log(x) - b*d**2*e*n*x**3/3 + b*d**2*e*x**3*log(c) + 3*b*d*e**2*n*x**4*
log(x)/4 - 3*b*d*e**2*n*x**4/16 + 3*b*d*e**2*x**4*log(c)/4 + b*e**3*n*x**5*log(x)/5 - b*e**3*n*x**5/25 + b*e**
3*x**5*log(c)/5

________________________________________________________________________________________

Giac [A]  time = 1.20664, size = 230, normalized size = 1.89 \begin{align*} \frac{1}{5} \, b n x^{5} e^{3} \log \left (x\right ) + \frac{3}{4} \, b d n x^{4} e^{2} \log \left (x\right ) + b d^{2} n x^{3} e \log \left (x\right ) - \frac{1}{25} \, b n x^{5} e^{3} - \frac{3}{16} \, b d n x^{4} e^{2} - \frac{1}{3} \, b d^{2} n x^{3} e + \frac{1}{5} \, b x^{5} e^{3} \log \left (c\right ) + \frac{3}{4} \, b d x^{4} e^{2} \log \left (c\right ) + b d^{2} x^{3} e \log \left (c\right ) + \frac{1}{2} \, b d^{3} n x^{2} \log \left (x\right ) - \frac{1}{4} \, b d^{3} n x^{2} + \frac{1}{5} \, a x^{5} e^{3} + \frac{3}{4} \, a d x^{4} e^{2} + a d^{2} x^{3} e + \frac{1}{2} \, b d^{3} x^{2} \log \left (c\right ) + \frac{1}{2} \, a d^{3} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)^3*(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

1/5*b*n*x^5*e^3*log(x) + 3/4*b*d*n*x^4*e^2*log(x) + b*d^2*n*x^3*e*log(x) - 1/25*b*n*x^5*e^3 - 3/16*b*d*n*x^4*e
^2 - 1/3*b*d^2*n*x^3*e + 1/5*b*x^5*e^3*log(c) + 3/4*b*d*x^4*e^2*log(c) + b*d^2*x^3*e*log(c) + 1/2*b*d^3*n*x^2*
log(x) - 1/4*b*d^3*n*x^2 + 1/5*a*x^5*e^3 + 3/4*a*d*x^4*e^2 + a*d^2*x^3*e + 1/2*b*d^3*x^2*log(c) + 1/2*a*d^3*x^
2